Question 122

If the $$n^{th}$$ term of the G.P. $$2, 2\sqrt{2}, 4, ......$$ is 128, then n =

Solution

Here a = 2

and r = $$\sqrt{2}$$

nth term = 128

$$T\ \left(n\right)\ =\ a.r^{\left(n-1\right)}$$

Putting the values,

128 =$$\ 2.\sqrt{2}^{\left(n-1\right)}$$

$$64=2^{\frac{\left(n-1\right)}{2}}$$

$$2^6=2^{\frac{\left(n-1\right)}{2}}$$

Comparing the exponents

$$6={\frac{\left(n-1\right)}{2}}$$

On solving,n = 13 Answer

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